- #1
Brandon O
- 2
- 0
Integral of unknown difficulty!
integral of: 1/[4+5sin(2x)] dx
U substitution on 2x gives u=2x
then du becomes 2dx so... du=2 dx
after that, i get
1/2 * int of: 1/[4+5sin(u)] du
from here, using v=5sin(u)
then: dv=5cos(u) du
this is where i get stuck. Did i start it off right or did I mess it up here? I don't need the answer to it, just a push in the right direction from the start ><
Homework Statement
integral of: 1/[4+5sin(2x)] dx
The Attempt at a Solution
U substitution on 2x gives u=2x
then du becomes 2dx so... du=2 dx
after that, i get
1/2 * int of: 1/[4+5sin(u)] du
from here, using v=5sin(u)
then: dv=5cos(u) du
this is where i get stuck. Did i start it off right or did I mess it up here? I don't need the answer to it, just a push in the right direction from the start ><